Question

a 0.152 kg baseball, thrown with a speed of 39.8 m/s, is hit straight back at the pitcher with a speed of 49.0 m/s.
(a) what is the magnitude of the impulse delivered by the bat to the baseball?
-13.5
x
net impulse is the change in momentum. since all other forces are negligible compared to the force of the bat, the impulse delivered by the bat is the net impulse. use momentum ideas to find it. kg·m/s
(b) find the magnitude of the average force exerted by the bat on the ball if the two are in contact for 1.90×10^{-3} s.
7.11
kn

Explanation:

Step1: Calculate the initial and final momentum

The formula for momentum is \(p = mv\).
Let the initial velocity \(v_{i}=39.8\ m/s\) and final velocity \(v_{f}=- 49.0\ m/s\) (negative because direction is reversed), and mass \(m = 0.152\ kg\).
Initial momentum \(p_{i}=mv_{i}=0.152\times39.8\ kg\cdot m/s\)

$$p_{i}=6.0496\ kg\cdot m/s$$

Final momentum \(p_{f}=mv_{f}=0.152\times(-49.0)\ kg\cdot m/s\)

$$p_{f}=-7.448\ kg\cdot m/s$$

Step2: Calculate the impulse (change in momentum)

Impulse \(J=\Delta p=p_{f}-p_{i}\)

$$J=-7.448 - 6.0496=-13.4976\ kg\cdot m/s$$

The magnitude of the impulse \(|J| = 13.5\ kg\cdot m/s\)

Step3: Calculate the average force (for part b)

We know that \(J = F_{avg}\Delta t\), where \(\Delta t = 1.90\times10^{-3}\ s\)
From \(F_{avg}=\frac{J}{\Delta t}\), substituting \(J = 13.4976\ kg\cdot m/s\)

$$F_{avg}=\frac{13.4976}{1.90\times 10^{-3}}\ N$$

$$F_{avg}=\frac{13.4976}{0.0019}\ N\approx7104\ N = 7.104\ kN\approx7.11\ kN$$

Answer:

(a) The magnitude of the impulse is \(13.5\ kg\cdot m/s\)
(b) The magnitude of the average force is \(7.11\ kN\)